a, PTHH:
\(\left(C_{17}H_{35}COO\right)_3C_3H_5+3NaOH\rightarrow3C_{17}H_{35}COONa\) \(+C_3H_5\left(OH\right)_3\)
b, Ta có: \(n_{\left(C_{17}H_{35}COO\right)_3C_3H_5}=\dfrac{8,9}{890}=0,01\left(mol\right)\)
Theo pt+đb,ta có: \(n_{C_3H_5\left(OH\right)_3}=n_{\left(C_{17}H_{35}COO\right)_3C_3H_5}=0,01\left(mol\right)\)
\(n_{C_{17}H_{35}COONa}=3.n_{\left(C_{17}H_{35}COO\right)_3C_3H_5}=3.0,01=0,03\left(mol\right)\)
=> m = \(m_{C_3H_5\left(OH\right)_3}=0,01.92=0,92\left(g\right)\)
Khối lượng muối tạo thành là:
\(m_{C_{17}H_{35}COONa}=0,03.306=9,18\left(g\right)\)
=.= hk tốt!!